EMM for Bachelier model

Question

The stock price is assumed to evolve as $S_{t}=S_{0}+\mu t+\sigma B_{t}$, where $S_{0}>0, \mu>0$ and the process $B_{t}$ is Brownian motion.

The saving account is assumed to be $\beta_{t}=e^{r t}$, with interest rate $r$

A call option with strike $K$ and expiration $T$ pays $C_{T}=\left(S_{T}-K\right)^{+}$ at time $T$.

Assume r = 0. Give the EMM.

My attempt

I am a bit lost when it comes to EMM but this is what I have so far:

• Girsanov's theorem:

$B_{t}$ is a B.M under measure P and C is a constant. Then there exists an EMM q such that:

$\hat{B}_{t}=B_{t}+C_{t} \sim Q$ brownian motion.

$d S_{t}=\mu d t+\sigma d B_{t}$

$r=0 \quad c=\frac{\mu-r}{\sigma}=c=\frac{\mu}{\sigma}$

I am not sure how to continue and give the EMM explicitly.

Edit: After some researching, I have found that the EMM does exist for Bachelier model and is unique by Girsanov's theorem. But I am still a bit lost on how to find it.

Answer 1

It sounds to me like you understand everything apart from how Girsanov's theorem defines the EMM.

Girsanov's theorem tells us that if $B_t$ is standard Brownian motion under $P$, then for any adapted process $\gamma_t$ (satisfying certain conditions) the process $\hat{B}_t$ defined by:

\begin{equation} d\hat{B}_t = \gamma_t dt +dB_t \end{equation} is Brownian motion under another equivalent measure and this equivalent measure (let's call it $Q$) can be defined by its Radon-Nikodym derivative:

\begin{equation} \frac{dQ_T}{dP} = \exp\bigg\{ -\int_0^T \gamma^\top(s) dB_s - \frac{1}{2} \int_0^T \gamma^2(s) ds \bigg\}. \end{equation}

Now, as you have already noticed, what I am calling $\gamma(t)$ should in our case be equal to $\frac{\mu - r}{\sigma} = \frac{\mu}{\sigma}$. Then $d\hat{B}_t = \frac{\mu}{\sigma} dt +dB_t$ and we have: \begin{equation} dS_t = \sigma d\hat{B}_t \end{equation} as desired. So, our EMM, $Q = Q_T \sim P$, is defined by the Radon-Nikodym derivative: \begin{align} \frac{dQ_T}{dP} &= \exp\bigg\{ -\int_0^T \frac{\mu}{\sigma} dB_s - \frac{1}{2} \int_0^T \frac{\mu^2}{\sigma^2} ds \bigg\} \\ &= \exp \Big\{ -\frac{\mu}{\sigma}B_T - \frac{1}{2} \frac{\mu^2}{\sigma^2}T \Big\} \end{align}